Skip to main content
Back to Height and Distance
JEE Main 2019
Height and Distance
Height and Distance
Easy

Question

ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot –1 (32\sqrt 2 ) and cosec –1 (22\sqrt 2 ) respectively, then the height of the tower (in metres) is :

Options

Solution

Δ\Delta APM hAM=132{h \over {AM}} = {1 \over {3\sqrt 2 }} Δ\Delta BPM hBM=17{h \over {BM}} = {1 \over {\sqrt 7 }} Δ\Delta ABM AM 2 + MB 2 = (100) 2 ' \Rightarrow 18h 2 + 7h 2 = 100 × 100 \Rightarrow h 2 = 4 × 100 \Rightarrow h = 20

Practice More Height and Distance Questions

View All Questions